# An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

**Question:**

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.

**Solution:**

Since the height and width of the arc from the centre is 2 m and 8 m respectively, it is clear that the length of the major axis is 8 m, while the length of the semi-minor axis is 2 m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the *x*-axis. Hence, the semi-ellipse can be diagrammatically represented as

The equation of the semi-ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, y \geq 0$, where $a$ is the semi-major axis

Accordingly, $2 a=8 \Rightarrow a=4$

*b* = 2

Therefore, the equation of the semi-ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1, y \geq 0$ (1)

Let A be a point on the major axis such that AB = 1.5 m.

Draw $A C \perp O B$.

OA = (4 – 1.5) m = 2.5 m

The *x*-coordinate of point C is 2.5.

On substituting the value of *x* with 2.5 in equation (1), we obtain

$\frac{(2.5)^{2}}{16}+\frac{y^{2}}{4}=1$

$\Rightarrow \frac{6.25}{16}+\frac{y^{2}}{4}=1$

$\Rightarrow y^{2}=4\left(1-\frac{6.25}{16}\right)$

$\Rightarrow y^{2}=4\left(\frac{9.75}{16}\right)$

$\Rightarrow y^{2}=2.4375$

$\Rightarrow y=1.56 \quad$ (approx.)

$\therefore A C=1.56 \mathrm{~m}$

Thus, the height of the arch at a point 1.5 m from one end is approximately 1.56 m.